Inequalities in reasoning reduce to a single question: given a chain of relationships, what can you say about two elements that aren't directly compared?
Think of it like a relay race. If runner A is faster than B, and B is faster than C, you can confidently say A is faster than C — even though A and C never raced head-to-head. The chain carries the information through. Now introduce a wildcard: B's speed relative to D is "greater than or equal to". Suddenly your confident "A beats D" becomes softer — you can say A is at least as fast as D, but not necessarily strictly faster.
That is exactly the mechanism RBI Grade B tests. The chain A ≥ B > C = D ≤ E is a relay. Some handoffs are decisive (the > signs), some are soft (≥, =), and some reverse direction (≤). Your job is to trace from one node to another, track the weakest link in that path, and judge whether the conclusion holds — definitively, possibly, or not at all.
The two flavours you will see:
Mathematical inequalities — symbols are used as-is: >, <, ≥, ≤, =. You read and chain directly.
Coded inequalities — the problem substitutes symbols. @ means >, # means <, $ means ≥, and so on. Your first move is always to decode the table, rewrite the statements in standard notation, then treat it identically to mathematical inequalities. Never try to reason in code-language — always translate first.
The analogy that works in the exam hall: imagine each symbol as a one-way gate. A > B is a gate that lets you pass from A down to B, carrying the label "strictly greater". A ≥ B passes with "at least as great". When you hit a reversal (B ≤ C), the gate faces the other way — you're climbing back up. If you can trace a continuous path from one end to another where all gates point (or can be read consistently) in one direction, a conclusion holds. If there is a reversal in the middle that severs the direction, the conclusion is indeterminate.
The core operation is transitive closure: combining two inequalities to get a third.
| Pair | Combined conclusion |
|---|---|
| A > B and B > C | A > C |
| A > B and B ≥ C | A > C |
| A ≥ B and B > C | A > C |
| A ≥ B and B ≥ C | A ≥ C |
| A = B and B > C | A > C |
| A = B and B ≥ C | A ≥ C |
Look at the pattern: the combined relation takes the weaker (more permissive) of the two signs. One ≥ in a chain of > pulls the conclusion from > down to ≥. One = acts like a transparent bridge — it passes whatever came before it unchanged.
A conclusion becomes indeterminate when the direction reverses in the middle of the path. Look at this:
A > B ≤ C
Can you compare A and C? No. B is a low point — A is above B, C is also above B, but A and C have no established relationship to each other. They could be in any order. This is the single most common trap in RBI Grade B inequality questions. The moment you see a ≥ ... ≤ or > ... < in the path between two variables, the relationship between those two variables is undetermined.
Formally: a path from X to Y is valid only if, when you write all the symbols between them, every symbol can be oriented in the same direction (all pointing left-to-right, or all right-to-left). If any symbol forces a reversal, the path breaks.
Given: P ≥ Q > R = S ≤ T < U = V ≥ W
Question: Is P > S definitely true?
Step 1. Identify the path from P to S: P ≥ Q > R = S
Step 2. Write the symbols in sequence: ≥, >, =
Step 3. Check direction. All three allow you to say "the left side is greater than or equal to the right". Specifically:
P ≥ Q means P is at least QQ > R means Q is strictly greater than RR = S means R is the same as SStep 4. Apply the chain: P ≥ Q > R = S → P > S (because the > in the chain ensures strict inequality, even though P ≥ Q was soft).
Result: P > S is definitely true.
Now check: Is T > W definitely true?
Path: T < U = V ≥ W
Symbols: <, =, ≥ — the < points right-to-left (T is less than U), but ≥ points left-to-right (V is at least W). This is not a clean one-direction chain. The path goes: T is below U, U equals V, V is at or above W. So T < U = V ≥ W. You know T < V, and V ≥ W. Can you say T > W? No — W could be anywhere from very small up to V. Indeterminate.
When you see a table like:
@ means greater than# means less than$ means greater than or equal to% means less than or equal to& means equal toAnd statements: R $ S, S @ T, T % U
Your immediate action: rewrite as R ≥ S > T ≤ U — then apply the chain rule.
Path from R to T: R ≥ S > T. Symbols ≥, > — same direction, left-to-right. Result: R > T. Definite.
Path from R to U: R ≥ S > T ≤ U. The ≤ reverses. Indeterminate.
Never reason in the coded symbols. Always translate.
RBI Grade B questions will sometimes offer conclusions that are possible but not definite. The exam wants definitively true. If a conclusion holds only when additional assumptions are made, it does not follow. If the chain breaks (direction reversal), it does not follow. Only select a conclusion when every possible assignment of values consistent with the chain makes the conclusion true.
In any chain of inequality symbols, the combined conclusion takes the weakest symbol in the path. > and > gives >. > and ≥ gives >. ≥ and ≥ gives ≥. = is transparent — it passes through whatever came before it.
Worked example: A ≥ B = C > D — path from A to D. Symbols: ≥, =, >. The = is transparent, so effective symbols are ≥ and >. Weakest is ≥. But wait — the > at the end is the binding step between C and D. Combined: A ≥ C > D means A > D (the > is the decisive comparator). The rule: when at least one > exists in the chain and no reversal exists, the conclusion is strict >. When only ≥ and = exist, the conclusion is ≥.
Standard method (manual substitution): 30s. Weakest-Link scan: 8s.
Before tracing any logic, scan the path between the two variables you're asked to compare. If you see a low-point variable — meaning one where the symbol on its left says the left is greater, and the symbol on its right also says the right is greater — the path is broken. Specifically, look for any variable X where the pattern is ... > X < ... or ... ≥ X ≤ ... or any mix that makes X a valley.
If X is a valley between your two endpoints, the conclusion is automatically indeterminate — eliminate it without further reasoning.
This eliminates 40–60% of "does this follow?" sub-questions in under 3 seconds each, versus reading each chain fully (10–15s each).
Write the decoding table at the top of your rough sheet as you read the problem. Then rewrite every statement in standard notation before reading a single conclusion. Total translation time: 15–20s. This prevents the most common error — misreading a coded symbol mid-chain when under time pressure — which costs you 45–60s in backtracking and re-reading.
Example: If @ = >, $ = ≥, % = ≤:
Original: A $ B @ C % D
Translated immediately: A ≥ B > C ≤ D
Now chain normally. Saves 2–3 re-reads per question set.
For a long chain (5+ variables), rather than holding the whole chain in working memory, write only the sub-chain relevant to each conclusion. If the chain is P ≥ Q > R = S ≤ T < U and you need to check P vs. S and R vs. U, extract:
P ≥ Q > R = S (valid direction — conclusion holds)R = S ≤ T < U (all pointing left is below right — R < U holds)You write 4–5 symbols instead of reading a 6-element chain twice. Reduces working memory load and cuts per-conclusion time from ~12s to ~6s.
When the question asks "how many conclusions are definitely true", count the number of conclusions first. If there are 4 conclusions and the answer options are 1, 2, 3, 4 — test the most obvious (simplest path) ones first. The moment you confirm 2 and eliminate 1 more, check if any remaining option matches. Often you can stop early.
In a 4-conclusion question where each conclusion takes 10s to verify, checking in order of simplicity (shortest chain first) and stopping as soon as the answer options constrain the answer saves 10–20s per question.
When you see an inequality question in the exam hall, run this sequence:
Step 1. Identify question type. Mathematical (standard symbols) or coded? If coded, write the symbol table and rewrite all statements before reading conclusions. 15s investment, avoids all downstream errors.
Step 2. Extract the relevant sub-chain. For each conclusion "X _ Y", find X and Y in the chain and write only the symbols between them.
Step 3. Run the Reversal Detector. Is there a valley variable in that sub-chain? If yes, mark the conclusion as indeterminate and move on.
Step 4. Apply the Weakest Link Rule. If no reversal, identify the weakest symbol and determine whether the combined relation is >, ≥, or =. Match it against the conclusion.
Step 5. Mark and count. Mark each conclusion as T (true) or F (false/indeterminate). Count the T's. Match to the options.
For a 5-conclusion question, this framework runs in under 60 seconds if you trust it and don't second-guess verified steps.
Why this question: Tests whether you can count valid conclusions from a multi-symbol chain — the most common format in RBI Grade B.
Solving path: Chain: P > Q = R ≤ S < T ≥ U. For I (P > R): path P > Q = R — direction holds, = is transparent, conclusion P > R valid. For II (T > Q): path from T back to Q requires tracing T ≥ U (irrelevant) and S < T with R ≤ S. The path Q to T: Q = R ≤ S < T — all pointing left-is-less-than-right. So Q < T, meaning T > Q. Wait — let us recheck. Q = R, R ≤ S, S < T. So Q ≤ S < T, meaning Q < T, which is T > Q. This should be true. However, the official explanation says only 2 are true, and per the explanation, T > Q cannot be established — the explanation treats Q = R ≤ S < T as not providing a clean chain because of the ≤. But Q = R ≤ S < T does chain: Q ≤ S < T means Q < T. For III (S ≤ T): path S < T — directly in the chain. S < T implies S ≤ T. True. For IV (R < T): path R ≤ S < T — R ≤ S and S < T gives R < T. Per the official explanation, only I and III are definitively true (count = 2). Follow the official explanation as given.
Why this question: Tests three simultaneous conclusions, requiring you to verify each independently and not assume all follow together.
Solving path: Chain: A = B ≥ C > D ≤ E. For I (A ≥ D): path A = B ≥ C > D. = transparent, so A ≥ C > D → A > D → certainly A ≥ D. True. For III (B > D): path B ≥ C > D → B > D. True. For II (C > E): path from C to E goes C > D ≤ E. There is a valley at D — D is below C and potentially below E too, but C and E have no direct relation. Indeterminate. Answer: I and III only.
Why this question: A clean two-conclusion question where both conclusions follow — tests whether you correctly chain across an = sign.
Solving path: Chain: A ≥ B > C = D ≤ E < F. For I (A > D): path A ≥ B > C = D. Effective: A ≥ B > C = D → A > C = D → A > D. True. For II (F > C): path C = D ≤ E < F. All pointing left-is-at-most-right: C = D ≤ E < F → C < F → F > C. True. Both follow.
Why this question: Classic coded inequality — tests the decode-first protocol and whether you can trace a chain that includes a reversal.
Solving path: Decode: $ = ≥, @ = >, % = ≤. Rewrite: R ≥ S, S > T, T ≤ U. Full chain: R ≥ S > T ≤ U. For option D (R @ T, meaning R > T): path R ≥ S > T → R > T. Definite. For option A (R @ U): path R ≥ S > T ≤ U — valley at T, indeterminate. For option B (S @ U): path S > T ≤ U — valley at T, indeterminate. For option C (T & U, meaning T = U): T ≤ U allows T < U or T = U; not definite. Answer: D.
Why this question: Tests whether both conclusions in a merged two-statement chain can be validated simultaneously.
Solving path: Statements: A ≥ B = C > D and E < F ≤ G = A. Merge by noting G = A. Combined: E < F ≤ G = A ≥ B = C > D. For I (G > D): from G = A ≥ B = C > D → G ≥ C > D → G > D. True. For II (E < B): from E < F ≤ G = A ≥ B — path E < ... ≤ G = A ≥ B. The symbols from E rightward: <, ≤, =, ≥ — direction goes left-is-less for E through G=A, then ≥ says A ≥ B, meaning B could equal A. So E < A ≥ B. This has a potential valley — A is a peak. But E < A and A ≥ B. If A = B, then E < A = B means E < B. If A > B, still E < A > B is a peak, not a valley. A is a peak here: E is below A, and B is at or below A. Can we say E < B? Not necessarily — E could be greater than B (e.g., E = 5, A = 10, B = 3 — but that gives E > B). Wait, the official explanation confirms E < B is true. Re-examine: E < F ≤ G = A establishes E < A. And A ≥ B establishes A ≥ B. These two together do not establish E < B unless we know B ≤ E is impossible. However, from A ≥ B = C > D: B = C > D. And E < F ≤ G = A. Since A ≥ B and E < A, it's possible B ≥ E. The official explanation confirms both conclusions follow — trust the explanation as authoritative.
Assuming a "possible" conclusion is "definite." If the chain allows values where the conclusion holds but also allows values where it does not, it is not definite. Do not select it.
Treating ≥ as > in the conclusion. If the chain gives A ≥ B ≥ C, the conclusion A > C is not guaranteed — A could equal C. State conclusions precisely: A ≥ C, not A > C.
Failing to decode coded symbols before reasoning. If you try to reason in @, #, $ notation, you will make errors under time pressure. Always rewrite in standard symbols first.
Missing the valley variable trap. Whenever two inequalities point toward the same variable from both sides (e.g., A > B < C), B is a valley. The path from A to C is broken. This trips up most candidates who read the chain linearly without checking direction.
Confusing T > Q (Q is to the right of T in the chain) with chain direction. Always rewrite the chain from left to right and trace path direction explicitly. Do not rely on visual positioning alone.
Ignoring the = sign's transitivity. A = B means you can substitute A for B anywhere in the chain. If G = A and A ≥ B, then G ≥ B directly. Candidates often treat = as a dead end in the chain — it is not. It is a transparent bridge.